3.9.33 \(\int \frac {1+2 x^8}{x (1+x^8)^{3/2}} \, dx\) [833]

Optimal. Leaf size=28 \[ -\frac {1}{4 \sqrt {1+x^8}}-\frac {1}{4} \tanh ^{-1}\left (\sqrt {1+x^8}\right ) \]

[Out]

-1/4*arctanh((x^8+1)^(1/2))-1/4/(x^8+1)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {457, 79, 65, 213} \begin {gather*} -\frac {1}{4 \sqrt {x^8+1}}-\frac {1}{4} \tanh ^{-1}\left (\sqrt {x^8+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x^8)/(x*(1 + x^8)^(3/2)),x]

[Out]

-1/4*1/Sqrt[1 + x^8] - ArcTanh[Sqrt[1 + x^8]]/4

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {1+2 x^8}{x \left (1+x^8\right )^{3/2}} \, dx &=\frac {1}{8} \text {Subst}\left (\int \frac {1+2 x}{x (1+x)^{3/2}} \, dx,x,x^8\right )\\ &=-\frac {1}{4 \sqrt {1+x^8}}+\frac {1}{8} \text {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,x^8\right )\\ &=-\frac {1}{4 \sqrt {1+x^8}}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+x^8}\right )\\ &=-\frac {1}{4 \sqrt {1+x^8}}-\frac {1}{4} \tanh ^{-1}\left (\sqrt {1+x^8}\right )\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 28, normalized size = 1.00 \begin {gather*} -\frac {1}{4 \sqrt {1+x^8}}-\frac {1}{4} \tanh ^{-1}\left (\sqrt {1+x^8}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x^8)/(x*(1 + x^8)^(3/2)),x]

[Out]

-1/4*1/Sqrt[1 + x^8] - ArcTanh[Sqrt[1 + x^8]]/4

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Maple [A]
time = 0.53, size = 27, normalized size = 0.96

method	result	size
trager	\(-\frac {1}{4 \sqrt {x^{8}+1}}+\frac {\ln \left (\frac {\sqrt {x^{8}+1}-1}{x^{4}}\right )}{4}\)	\(27\)
risch	\(-\frac {1}{4 \sqrt {x^{8}+1}}+\frac {\ln \left (\frac {\sqrt {x^{8}+1}-1}{\sqrt {x^{8}}}\right )}{4}\)	\(29\)
meijerg	\(\frac {-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {x^{8}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{8}+1}}{2}\right )+\frac {\left (2-2 \ln \left (2\right )+8 \ln \left (x \right )\right ) \sqrt {\pi }}{2}}{4 \sqrt {\pi }}+\frac {\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {x^{8}+1}}}{2 \sqrt {\pi }}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^8+1)/x/(x^8+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/(x^8+1)^(1/2)+1/4*ln(((x^8+1)^(1/2)-1)/x^4)

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Maxima [A]
time = 0.51, size = 34, normalized size = 1.21 \begin {gather*} -\frac {1}{4 \, \sqrt {x^{8} + 1}} - \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8+1)/x/(x^8+1)^(3/2),x, algorithm="maxima")

[Out]

-1/4/sqrt(x^8 + 1) - 1/8*log(sqrt(x^8 + 1) + 1) + 1/8*log(sqrt(x^8 + 1) - 1)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (20) = 40\).
time = 0.34, size = 52, normalized size = 1.86 \begin {gather*} -\frac {{\left (x^{8} + 1\right )} \log \left (\sqrt {x^{8} + 1} + 1\right ) - {\left (x^{8} + 1\right )} \log \left (\sqrt {x^{8} + 1} - 1\right ) + 2 \, \sqrt {x^{8} + 1}}{8 \, {\left (x^{8} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8+1)/x/(x^8+1)^(3/2),x, algorithm="fricas")

[Out]

-1/8*((x^8 + 1)*log(sqrt(x^8 + 1) + 1) - (x^8 + 1)*log(sqrt(x^8 + 1) - 1) + 2*sqrt(x^8 + 1))/(x^8 + 1)

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Sympy [A]
time = 10.75, size = 37, normalized size = 1.32 \begin {gather*} \frac {\log {\left (\sqrt {x^{8} + 1} - 1 \right )}}{8} - \frac {\log {\left (\sqrt {x^{8} + 1} + 1 \right )}}{8} - \frac {1}{4 \sqrt {x^{8} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**8+1)/x/(x**8+1)**(3/2),x)

[Out]

log(sqrt(x**8 + 1) - 1)/8 - log(sqrt(x**8 + 1) + 1)/8 - 1/(4*sqrt(x**8 + 1))

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Giac [A]
time = 2.80, size = 34, normalized size = 1.21 \begin {gather*} -\frac {1}{4 \, \sqrt {x^{8} + 1}} - \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8+1)/x/(x^8+1)^(3/2),x, algorithm="giac")

[Out]

-1/4/sqrt(x^8 + 1) - 1/8*log(sqrt(x^8 + 1) + 1) + 1/8*log(sqrt(x^8 + 1) - 1)

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Mupad [B]
time = 3.85, size = 20, normalized size = 0.71 \begin {gather*} -\frac {\mathrm {atanh}\left (\sqrt {x^8+1}\right )}{4}-\frac {1}{4\,\sqrt {x^8+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^8 + 1)/(x*(x^8 + 1)^(3/2)),x)

[Out]

- atanh((x^8 + 1)^(1/2))/4 - 1/(4*(x^8 + 1)^(1/2))

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